Inequality of Incomplete Sum for $e^n$ Versus $e^{n}/2$

Question

Several years ago, I encountered a problem:

Prove that, for all natural numbers $n\ge 1$ we have

$$\sum_{k=0}^{n-1} \frac{n^k}{k!}<\frac{e^n}{2}<\sum_{k=0}^{n} \frac{n^k}{k!}.$$

The original solution is using induction, but that solution is false. Recently I am revisiting the problems and I found this one, so I am asking here.

P.S. I have read the post how to ask a good question. I really don't know how to go on with this problem. Please don't regard this question as no-clue questions, thanks. To answer the commonly asked question "what attempts have I tried." I would like to say that I have tried to use a different induction, but it gives no result. Also, by graphing, the solution of $\frac{e^x}{2}=\sum_{k=0}^{n} \frac{x^k}{k!}$ I have observed is around $n+\frac 23$ (I know that it is not allowed to ask two questions in one post, but I will put this observation here and if this question is proved, I am going to ask a separate question for this observation.)

Answer 1

Repeated integration by parts applied to the incomplete gamma function $$ \Gamma (n + 1,z) = \int_z^{ + \infty } {t^n e^{ - t} dt} $$ yields $$ \sum\limits_{k = 0}^n {\frac{{z^k }}{{k!}}} = e^z \frac{{\Gamma (n + 1,z)}}{{\Gamma (n + 1)}}. $$ Thus, the problem is equaivalent to $$ \frac{{\Gamma (n,n)}}{{\Gamma (n)}} < \frac{1}{2} < \frac{{\Gamma (n + 1,n)}}{{\Gamma (n + 1)}}. $$ Now using the known recurrence relation $$ \frac{{\Gamma (n,n)}}{{\Gamma (n)}} - \frac{{\Gamma (n + 1,n + 1)}}{{\Gamma (n + 1)}} = \frac{{\int_n^{n + 1} {t^n e^{ - t} dt} - n^n e^{ - n} }}{{\Gamma (n + 1)}} < 0 $$ and $$ \frac{{\Gamma (n + 1,n)}}{{\Gamma (n + 1)}} - \frac{{\Gamma (n,n - 1)}}{{\Gamma (n)}} = \frac{{(n - 1)^n e^{ - (n - 1)} - \int_{n - 1}^n {t^n e^{ - t} dt} }}{{\Gamma (n + 1)}} < 0, $$ since the integrand has a peak at $t=n$. Therefore, $$ \frac{{\Gamma (n,n)}}{{\Gamma (n)}} $$ is strictly monotonically increasing and $$ \frac{{\Gamma (n + 1,n)}}{{\Gamma (n + 1)}} $$ is strictly monotonically decreasing. It is enough to show that both sequences converge to $\frac{1}{2}$. For any fixed $a$, we have by Laplace's method $$ \Gamma (n + a,n) = \int_n^{ + \infty } {t^{n + a - 1} e^{ - t} dt} = n^{n + a} e^{ - n} \int_0^{ + \infty } {e^{ - n(e^s - s - 1)} e^{as} dt} \sim n^{n + a - 1/2} e^{ - n} \sqrt {\frac{\pi }{2}} $$ as $n\to +\infty$. Employing this result in the special cases $a=0,1$ together with Stirling's formula $$ \Gamma (n) \sim n^{n - 1/2} e^{ - n} \sqrt {2\pi } ,\quad \Gamma (n + 1) = n\Gamma (n) \sim n^{n + 1/2} e^{ - n} \sqrt {2\pi } $$ yields the desired limit.

Note. One of the inequalities can be proved without Laplace's method as follows. We observe that \begin{align*} \frac{{\Gamma (n,n)}}{{\Gamma (n)}} & = \frac{{\int_0^{ + \infty } {e^{ - n(e^s - s )} ds} }}{{\int_{ - \infty }^{ + \infty } {e^{ - n(e^s - s )} ds} }} = \left( {1 + \frac{{\int_{ - \infty }^0 {e^{ - n(e^s - s )} ds} }}{{\int_0^{ + \infty } {e^{ - n(e^s - s )} ds} }}} \right)^{ - 1} \\ & = \left( {1 + \frac{{\int_0^{ + \infty } {e^{ - n(e^{ - s} + s )} ds} }}{{\int_0^{ + \infty } {e^{ - n(e^s - s )} ds} }}} \right)^{ - 1} . \end{align*} Now $e^{ - s} + s < e^s - s \Longleftrightarrow s < \sinh s$ for $s>0$, whence the ratio of the two integrals is $>1$.

Answer 2

Here's a proof from a probabilistic perspective, adapted from the lemmas in [1]. The statement is equivalent to proving \begin{align*} P(N_n \le n-1) \overset{(1)}{<} \frac{1}{2} \overset{(2)}{<} P(N_n \le n) \end{align*}

Proof of (2)

Let $N_\lambda \sim \text{Poisson}(\lambda)$ and define $g_n(u) := e^{-u} u^n$. Then we have \begin{align*} P(N_n \le n) - P(N_{n+1} \le n+1) &= \color{green}{P(N_n \le n+1) - P(N_{n+1} \le n+1))} - \color{orange}{P(N_n = n+1)} \\ &= \color{green}{\frac{1}{(n+1)!}\int_{n}^{n+1}g_{n+1}(u) du} - \color{orange}{\frac{1}{(n+1)!}\int_{n}^{n+1}g_{n+1}(n) du} \\ &= \frac{1}{(n+1)!}\int_{n}^{n+1}[g_{n+1}(u) - g_{n+1}(n)]du \\ &>0 \end{align*} with the final inequality arising from the fact that $g_{n+1}(u)$ is increasing on $[n, n+1]$. This means that $$a_n := P(N_n \le n)$$ is a monotone-decreasing sequence. But by the Central Limit Theorem, $$P(N_n \le n) = \Phi(0) + o(1) \rightarrow \frac{1}{2}$$ as $n \rightarrow \infty$. Therefore, $a_n > \frac{1}{2}$ for all $n \ge 1$.

Proof of (1) A similar proof from above also shows that $P(N_n \ge n)$ also monotonically decreases to $\frac{1}{2}$. Therefore, \begin{align*} P(N_n \le n-1) = 1 - P(N_n \ge n) < \frac{1}{2} \end{align*} for all $n$.

[1] Adell, J. A.; Jodrá, P., The median of the Poisson distribution, Metrika 61, No. 3, 337-346 (2005). ZBL1079.62014.

Answer 3

My attempt :

We introduce the function :

$$g(x)=-\sum_{k=0}^{n-1}\frac{x^{\left(k\right)}}{k!}+\frac{e^{x}}{2}$$

Using strong induction

you can show that the function $g(x)$ is increasing and convex around $x=n$ or $x\in [n,n+1]$.

Next we can use the inequality : $$f(x)\geq f'(n)(x-n)+f(n)+0.5f''(x)(x-n)^2$$

Now we use $x=n+1$ .

On the other hand we have for $n\ge 13$: $$\frac{2(n+1)^n}{3n!}-\frac{3n^{(n-1)}}{2(n-1)!}-\frac{n^{n-2}}{2(n-2)!}\le 0$$

To be continued.

edit 06/02/2022:

We introduce the function :

$$f(x)=\frac{e^{x}}{2}-\sum_{k=0}^{n-1}\frac{x^{k}}{k!}$$

Using Hermite Hadamard inequality on $[n,n+1]$ ($f(x)$ is convex on this interval ) we get the inequality :

$$\frac{e^{n+1}}{2}-\sum_{k=0}^{n-1}\frac{\left(n+1\right)^{k}}{k!}+\frac{e^{n}}{2}-\sum_{k=0}^{n-1}\frac{n^{k}}{k!}\geq e^{n+1}-2\sum_{k=0}^{n}\frac{\left(n+1\right)^{k}}{k!}-\left(e^{n}-2\sum_{k=0}^{n}\frac{n^{k}}{k!}\right)$$

Now we can use the inequality for $x,i>1$ i a natural number :

$$e^x>1+x+\cdots+\frac{x^i}{i!}$$

We get the inequality :

$$\frac{e^{n+1}}{2}-\sum_{k=0}^{n-1}\frac{\left(n+1\right)^{k}}{k!}+\frac{e^{n}}{2}-\sum_{k=0}^{n-1}\frac{n^{k}}{k!}\geq \sum_{k=0}^{2n}\frac{\left(n+1\right)^{k}}{k!}-2\sum_{k=0}^{n}\frac{\left(n+1\right)^{k}}{k!}-\left(e^{n}-2\sum_{k=0}^{n}\frac{n^{k}}{k!}\right)$$

Or :

$$\frac{e^{n+1}}{2}-\sum_{k=0}^{n-1}\frac{\left(n+1\right)^{k}}{k!}+\frac{e^{n}}{2}-\sum_{k=0}^{n-1}\frac{n^{k}}{k!}\geq -\sum_{k=0}^{n}\frac{\left(n+1\right)^{k}}{k!}+\sum_{k=n+1}^{2n}\frac{\left(n+1\right)^{k}}{k!}-\left(e^{n}-2\sum_{k=0}^{n}\frac{n^{k}}{k!}\right)$$

Or :

$$\frac{e^{n+1}}{2}-\sum_{k=0}^{n-1}\frac{\left(n+1\right)^{k}}{k!}\geq -\frac{n^{n}}{n!}-\sum_{k=0}^{n}\frac{\left(n+1\right)^{k}}{k!}+\sum_{k=n+1}^{2n}\frac{\left(n+1\right)^{k}}{k!}-\left(\frac{3}{2}e^{n}-3\sum_{k=0}^{n}\frac{n^{k}}{k!}\right)$$

Now I cannot show it but a numerical experiment shows that for $n\geq 7$ we have :

$$-\frac{n^{n}}{n!}-\sum_{k=0}^{n}\frac{\left(n+1\right)^{k}}{k!}+\sum_{k=n+1}^{2n}\frac{\left(n+1\right)^{k}}{k!}>0$$

And already proved here Prove this inequality $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}>\frac{e^x}{2}$ we have :

$$-\left(\frac{3}{2}e^{n}-3\sum_{k=0}^{n}\frac{n^{k}}{k!}\right)>0$$

It shows a weaker inequality but we may consider some improvement .

Well gives me feedback and feel free to make comment .

Last Edit 07/02/2022 :

By improvement we can use the theorem 1 here https://www.researchgate.net/publication/329884875_Hermite-Hadamard_inequality_for_M_ph_A-strongly_convex_functions

We have using the theorem 1 the inequality :

$$\frac{n^{n-2}}{6\left(n-2\right)!}+\frac{n^{n-1}}{6\left(n-1\right)!}+\frac{1}{6}\frac{n^{n}}{n!}-0.15\frac{\left(n+1\right)^{\left(n+1\right)}}{\left(n+1\right)!}>0$$

And for $n\geq 13$ it seems we have :

$$-\frac{n^{n}}{n!}-\sum_{k=0}^{n}\frac{\left(n+1\right)^{k}}{k!}+\sum_{k=n+1}^{2n}\frac{\left(n+1\right)^{k}}{k!}-0.25\frac{\left(n+1\right)^{\left(n+1\right)}}{\left(n+1\right)!}>0$$

And it seems we have for $n\geq 4$ :

$$-\left(\left(\frac{3}{2}-\frac{1}{6}\right)e^{n}-\left(3-\frac{1}{3}\right)\sum_{k=0}^{n}\frac{n^{k}}{k!}\right)-0.6\frac{\left(n+1\right)^{\left(n+1\right)}}{\left(n+1\right)!}>0\tag{I}$$

Edit 15/02/2022

To show $(I)$ we can use the fact wich follows for $n\geq 10$ $$e^n-\sum_{k=0}^{n^2}\frac{n^k}{k!}<1$$

Reference :

Turhan, Sercan & Maden, Selahattin & Demirel, Ayse Kubra & İşcan, İmdat. (2018). Hermite-Hadamard inequality for M ϕ A-strongly convex functions.